Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const size_t MAX_N = 1e4 + 5;
vector<pair<int, int> > res;
int a[MAX_N], b[MAX_N], n;
inline void add(int *a, int i, int j) {
  a[i] ^= a[j];
  res.emplace_back(i + 1, j + 1);
}
void work(int *a) {
  for (int i = 30, st = 0; i >= 0; --i)
    for (int j = st; j < n; ++j)
      if (a[j] >> i & 1) {
        if (j != st) {
          add(a, j, st);
          add(a, st, j);
          add(a, j, st);
        }
        for (int k = 0; k < n; ++k)
          if (k != st && (a[k] >> i & 1)) add(a, k, st);
        ++st;
        break;
      }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", a + i);
  for (int i = 0; i < n; ++i) scanf("%d", b + i);
  vector<pair<int, int> > resa, resb;
  work(a);
  resa.swap(res);
  work(b);
  resb.swap(res);
  res.swap(resa);
  for (int i = 0; i < n; ++i)
    if (a[i]) {
      int bit = 30;
      while (!(a[i] >> bit & 1)) --bit;
      for (int j = 0; j <= 30 && j < n; ++j)
        if ((a[j] ^ b[j]) >> bit & 1) add(a, j, i);
    }
  res.insert(res.end(), resb.rbegin(), resb.rend());
  for (int i = 0; i < n; ++i)
    if (a[i] != b[i]) return puts("-1"), 0;
  printf("%lu\n", res.size());
  for (pair<int, int> sol : res) printf("%d %d\n", sol.first, sol.second);
  return 0;
}